What is the slope of the tangent line of # x^2/(x-y^3) = C #, where C is an arbitrary constant, at #(-2,1)#?

Question

1188 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-02T08:02:11

#-3/2#

Find the implicit derivative of the function.

#((x-y^3)d/dx(x^2)-x^2d/dx(x-y^3))/(x-y^3)^2=0#

Remember that differentiating a #y# term will spit out a #dy/dx# term thanks to the chain rule.

#(2x(x-y^3)-x^2(1-3y^2dy/dx))/(x-y^3)^2=0#

Plug in #(-2,1)# for #(x,y)# and solve for #dy/dx#.

#(2(-2)(-2-1^3)-(-2)^2(1-3(1)^2dy/dx))/(-2-1^3)^2=0#

Notice that the denominator can be cancelled out. Continue simplification of numerator.

#(-4)(-3)-(4)(-2)dy/dx=0#

#12+8dy/dx=0#

#dy/dx=-3/2#

The slope of the tangent line is #-3/2#.